Re: Easy probability question (Bayes theorem)
- From: Ray Vickson <RGVickson@xxxxxxx>
- Date: Sun, 9 Dec 2007 21:59:48 -0800 (PST)
On Dec 9, 8:51 pm, Taras_96 <taras...@xxxxxxxxx> wrote:
Hi everyone,
I'm having a bit of trouble with probability and ordering. I'll try to
explain with an example:
QUESTION: what is the probability that you choose an ace of spades and
an ace of diamonds out of a deck of cards
Using Baye's theorem:
Pr(pick an ace of space and an ace of diamonds) = Pr(ace of
spaces)*Pr(ace of diamonds|ace of spades)
= 1/52 * 1/51
This seems to have a kind of order 'built in', that the solution is
only valid if you pick an ace of spades first, and THEN an ace of
diamonds.
Letting AD = "pick Ace of Diamonds" and AC = "... Spades", then P{AC &
AD} = P{AC then AD} + P{AD then AC} = P{AC}*P{AD|AC} + P{AD}*P{AC|AD}
= 2*1/52/51.
Using a slightly different tact, there is only one way you can choose
an ace of spaces and an ace of diamonds out of a pack, thus
Pr(pick an ace of space and an ace of diamonds) = 1/(52C2) =
2*(1/52*1/51), which implies in the first method we should have also
considered picking diamonds first and then spades. However, this
consideration of order is not apparent in Baye's theorem.
Actually, it is, because in applying Bayes' (NOT Baye's) methods you
need one event to be a "prior" and the other a "posterior".
How about a more complicated question?
Pr(pick two aces and a non-ace)
= pr(picking an ace)*(pr picking an ace| an ace has been
picked)*pr(picking not an ace | two aces have been picked)
but also you could have
= pr(picking a non-ace)*pr(pr picking an ace| non ace has been
picked)*pr(picking a an ace | non ace & ace has been picked)
....
Do you have to sum the probabilities for each 'order'? I don't
remember ever seeing this problem crop up when I was doing
probability..
If all you care about is the number of aces and non-aces, with no
regard for order, then YES, you need to sum over the different orders:
{2A & 1N} = {AAN} + {ANA} + {NAA} in that order. P{AAN} =
P{A}*P{another A|A}*P{N|2A} = (4/52)*(3/51)*(48/40), etc.
An alternative method: consider cards to be of two types: aces (A)
(number = 4) and non-aces (N) (number = 48). Given that you draw
exactly three cards at random without replacement, you want the
probability of getting exactly 2 aces (the other is automatically an
N). This is given by a hypergeometric distribution: in a population of
two types (4 As and 48 Ns), what is the probability that in a sample
of size 3 you get two of type A? See http://en.wikipedia.org/wiki/Hypergeometric_distribution
or http://mathworld.wolfram.com/HypergeometricDistribution.html for
appropriate formulas.
R.G. Vickson
.
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